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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 3

Graph each inequality. x−2y>10

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1
Rewrite the inequality \(x - 2y > 10\) to express \(y\) in terms of \(x\). Start by isolating the \(y\) term: subtract \(x\) from both sides to get \(-2y > 10 - x\).
Next, divide both sides of the inequality by \(-2\) to solve for \(y\). Remember that dividing by a negative number reverses the inequality sign, so you get \(y < \frac{10 - x}{-2}\).
Simplify the right side of the inequality: \(y < \frac{10}{-2} - \frac{x}{-2}\), which simplifies to \(y < -5 + \frac{x}{2}\) or \(y < \frac{x}{2} - 5\).
Graph the boundary line \(y = \frac{x}{2} - 5\) on the coordinate plane. Since the inequality is strict ($<$), use a dashed line to indicate that points on the line are not included in the solution.
Determine which side of the boundary line to shade by choosing a test point not on the line (commonly \((0,0)\)). Substitute into the inequality \(y < \frac{x}{2} - 5\) and check if the inequality holds. Shade the region where the inequality is true.

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Graphing Linear Inequalities

Graphing linear inequalities involves first graphing the related linear equation as a boundary line. The inequality determines which side of the line is shaded, representing all solutions that satisfy the inequality. For example, if the inequality is 'greater than,' the region above the line is shaded.
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Converting Inequalities to Slope-Intercept Form

To graph the inequality, rewrite it in slope-intercept form (y = mx + b) by isolating y. This makes it easier to identify the slope and y-intercept, which are essential for drawing the boundary line accurately on the coordinate plane.
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Using Test Points to Determine Shading

After graphing the boundary line, select a test point not on the line (commonly (0,0)) to check if it satisfies the inequality. If it does, shade the region containing that point; if not, shade the opposite side. This ensures the correct solution region is represented.
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